c - Clarification about precedence of operators

Question

I got this snippet from some exercises and the question: which is the output of following code:

main()
{
    char *p = "ayqm";
    printf("%c", ++*(p++));
}

My expected answer was z but the actual answer was in fact b. How is that possible?

Later edit: the snippet is taken as it is from an exercise and did not focus on the string literal or syntax issues existent in other than the printf() code zone.

标签： coperator-precedence

Answer 1

Your program is having undefined behavior because it is trying to modify the string literal "ayqm". As per the standard attempting to modify a string literal results in undefined behavior because it may be stored in read-only storage.

The pointer p is pointing to string literal "ayqm". This expression

printf ("%c", ++*(p++));

end up attempting to modify the string literal that pointer p is pointing to.

An undefined behavior in a program includes it may execute incorrectly (either crashing or silently generating incorrect results), or it may fortuitously do exactly what the programmer intended.